Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10, maxlog = 18;
vector<int> v[maxn];
vector<int> a[maxn];
int depth[maxn];
vector<int> s[maxlog][maxn];
int sparce[maxlog][maxn];
void dfs(int x, int p = 0) {
  depth[x] = depth[p] + 1;
  sparce[0][x] = p;
  for (int i : a[x]) {
    if (s[0][x].size() == 11) break;
    s[0][x].push_back(i);
  }
  for (int i : v[x]) {
    if (i == p) continue;
    dfs(i, x);
  }
}
int equalize(int x, int y) {
  for (int i = maxlog - 1; i >= 0; --i)
    if (depth[sparce[i][x]] >= depth[y]) x = sparce[i][x];
  return x;
}
int find_lca(int x, int y) {
  if (x == y) return x;
  for (int i = maxlog - 1; i >= 0; --i) {
    if (sparce[i][x] != sparce[i][y]) {
      x = sparce[i][x];
      y = sparce[i][y];
    }
  }
  return sparce[0][x];
}
int lca(int x, int y) {
  if (depth[x] > depth[y]) swap(x, y);
  return find_lca(x, equalize(y, x));
}
int mn;
vector<int> res;
bool print;
void merge(const vector<int> &a1, const vector<int> &b1) {
  int it_a = 0;
  int it_b = 0;
  res.clear();
  mn = 0;
  while (res.size() < 11 && (it_a < a1.size() || it_b < b1.size())) {
    while (it_b < b1.size() && it_a < a1.size() && a1[it_a] == b1[it_b]) ++it_b;
    while (!res.empty() && it_b < b1.size() && b1[it_b] == res.back()) ++it_b;
    while (!res.empty() && it_a < a1.size() && a1[it_a] == res.back()) ++it_a;
    if (it_b == b1.size()) {
      res.push_back(a1[it_a]);
      ++it_a;
      continue;
    }
    if (it_a == a1.size()) {
      res.push_back(b1[it_b]);
      ++it_b;
      continue;
    }
    if (b1[it_b] < a1[it_a]) {
      res.push_back(b1[it_b]);
      ++it_b;
    } else {
      res.push_back(a1[it_a]);
      ++it_a;
    }
  }
}
void f(int x, int lca, vector<int> &curr) {
  for (int i = maxlog - 1; i >= 0; --i) {
    if (depth[sparce[i][x]] >= depth[lca]) {
      merge(curr, s[i][x]);
      curr = res;
      x = sparce[i][x];
    }
  }
  merge(curr, {});
  curr = res;
}
vector<int> a1, b1, ans;
int x, y, z, n, m, t, len, curr_lca;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> m >> t;
  for (int i = 1; i < n; ++i) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= m; ++i) {
    cin >> x;
    a[x].push_back(i);
  }
  dfs(1);
  for (int i = 1; i < maxlog; ++i) {
    for (int j = 1; j <= n; ++j) {
      sparce[i][j] = sparce[i - 1][sparce[i - 1][j]];
      s[i][j] = s[i - 1][j];
      merge(s[i][j], s[i - 1][sparce[i - 1][j]]);
      s[i][j] = res;
    }
  }
  for (int i = 1; i <= n; ++i) {
    v[i].clear();
    a[i].clear();
  }
  for (int i = 0; i < t; ++i) {
    cin >> x >> y >> z;
    curr_lca = sparce[0][lca(x, y)];
    a1.clear();
    b1.clear();
    f(x, curr_lca, a1);
    f(y, curr_lca, b1);
    merge(a1, b1);
    a1 = res;
    if (a1.empty()) {
      cout << 0 << '\n';
      continue;
    }
    ans.clear();
    ans.push_back(a1[0]);
    for (int i = 1; i < a1.size(); ++i)
      if (a1[i] != a1[i - 1]) ans.push_back(a1[i]);
    len = min(z, (int)ans.size());
    cout << len << ' ';
    for (int i = 0; i < len; ++i) cout << ans[i] << ' ';
    cout << '\n';
  }
  return 0;
}